Problem: Which of the following numbers is a multiple of 3? ${43,103,106,114,119}$
Solution: The multiples of $3$ are $3$ $6$ $9$ $12$ ..... In general, any number that leaves no remainder when divided by $3$ is considered a multiple of $3$ We can start by dividing each of our answer choices by $3$ $43 \div 3 = 14\text{ R }1$ $103 \div 3 = 34\text{ R }1$ $106 \div 3 = 35\text{ R }1$ $114 \div 3 = 38$ $119 \div 3 = 39\text{ R }2$ The only answer choice that leaves no remainder after the division is $114$ $ 38$ $3$ $114$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $3$ are contained within the prime factors of $114$ $114 = 2\times3\times19 3 = 3$ Therefore the only multiple of $3$ out of our choices is $114$. We can say that $114$ is divisible by $3$.